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• Level: GCSE
• Subject: Maths
• Word count: 2099

# An investigation for working out hidden faces as different number of cubes are joined by making different shapes.

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Introduction

Amit Patel

Mathematics Coursework

An investigation for working out hidden faces as different number of cubes are joined by making different shapes.

As part of my GCSE mathematics requirements  I have been assigned to investigate the number of hidden faces as cubes are joined in various way I shall start by making diagrams on the dotted paper provided and shall work out an expression (formula) which will reflect a relationship between the hidden faces and the number of cubes used.

Part 1: Number of cubes

• When I take a single cube and place it on dotted paper I can see 1 face hidden and faces visible from total of 6 faces.
• Similarly when I join 2 cubes and place them flat on a surface I can figure out that 4 faces are hidden and 8 are visible from a total of 12 cubes.
• When 3 cubes are joined to a similar pattern as mentioned above a total of 7 hidden faces and 11 visible faces out of a possible 18 faces observed

I am presenting a small table which describes the number of cubes used, the hidden faces, the visible faces and the total number of faces

 nth term No of cubes No of hidden faces No of visible faces No of total faces 1 1 1 5 6 2 2 4 8 12 3 3 7 11 18 4 4 10 14 24 5 5 13 17 30 6 6 16 20 36 7 7 19 23 42

From the above table a simple sequence can be formed and an nth term of the sequence can be worked out

Sequence for hidden faces

1     4     7     10     13     16     19

3       3    3      3        3       3

By taking the 1st

Middle

13    13      13     13     13      13

Sequence for visible faces

11     16     21     26     31     36     41     the nth term for the sequence is 5n+11

5        5        5       5       5      5

• When I join 5 cubes into 4 rows and lay it onto a flat surface I can find a total number of 120 faces with 38 of those showing and 82 hidden. The following table shows the sequence of 4 cubes.
 nth term No of cubes No of hidden faces No of visible faces No of total faces 1 4 10 14 24 2 8 28 20 48 3 12 46 26 72 4 16 64 32 96 5 20 82 38 120 6 24 100 44 144 7 28 118 50 168

From the table 2 simple sequences can be formed and an nth term of the sequence

Can be worked out

Sequence for hidden faces

10     28     46     64     82     100     118     the nth term for the sequence is 18n-8

18     18    18     18      18       18

Sequence for visible faces

14     20     26     31     38     44     50     the nth term for the sequence is 6n+8

6       6        6        6      6       6

• When five cubes are joined together in 5 rows I can observe 105 hidden faces, 45 showing faces with a total of 150 faces all together.

This table will show the next seven terms of the sequence

 nth term No of cubes No of hidden faces No of visible faces No of total faces 1 5 13 17 30 2 10 36 24 60 3 15 59 31 90 4 20 82 38 120 5 25 105 45 150 6 30 128 52 180 7 35 151 59 210

From the table 2 simple sequences can be formed and an nth term of the sequence

Can be worked out

Sequence for hidden faces

13     36     59     82     59     82     105    128     151     the nth term for the sequence is

23n-10

23     23     23     23     23     23       23       23

Sequence for visible faces

17     24     31     38     45     52     59                            the nth term for the sequence is

7n+10

7         7       7       7       7       7

I have made this formula based on the above sequences which I have made by taking the height, width and the length of cuboids.

6 HWL-2HL-2HW-LW

Conclusion

No of total faces

1

9

27

27

54

2

18

72

36

108

3

27

117

45

162

4

36

162

54

216

5

45

207

63

270

6

54

252

72

324

7

63

297

81

378

From the above table a simple sequences can be formed and an nth term of the sequence

Can be worked out

Sequence for hidden faces

27     72     117     162     207     252     297 the nth term for the sequence 45n-18

45     45        45       45        45      45

Testifying the above sequences

Considering the formula used previously I am going to test the nth term made for bigger cuboids or cubes made from small cubes.

6 HWL-(LW+2HL+2HW)

1.   If I take the nth term of a cuboid which I made previously as 7n-4 and test it for the fifth term then it is equal to 31. By using the general formula L = 5 H = 2 W = 1.

6 x 2 x 1 x 5 - (5 x 1 + 2 x 5 x 2 + 2 x 2 x 1)

60-29=31

2.   By applying the nth term 18n-8 for the fifth term the answer comes to

18 x 5 - 8 = 82

By applying the final formula L = 5 H = 2 W = 2

6 x 5 x 2 x 2 - (5 x 2 + 2 x 2 x 5 + 2 x 2 x 2)

120 – 38 = 82

3. For my final example of a cuboid with the nth term of 45n-18 the value of the fifth term is 207 where as by applying the final formula for cuboids L = 5 H = 3 W = 3

6 x 5 x 3 x 3 - (5 x 3 + 2 x 3 x 5 + 2 x 3 x 3)

270 - 63 = 207

Evaluation

I conclude from various testing which I have conducted earlier that as the columns, rows and depth of a cuboid is altered the general formula produced will adjust itself accordingly and a true number of hidden faces can be worked out easily. This also explains the generalisations, predictions and other features considered during this assignment and strengthen the solution of hidden faces.

This student written piece of work is one of many that can be found in our GCSE Hidden Faces and Cubes section.

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